Coupled mass to matrix problem

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• Nov 30th 2009, 09:42 PM
Mollier
Coupled mass to matrix problem
Hi, I am new around here. Hope you don't mind me beginning my stay with a question.

1. The Problem

Suppose masses $m_{1}, m_{2}, m_{3}, m_{4}$ are located at positions $x_{1}, x_{2}, x_{3}, x_{4}$ in a line and connected by springs with constants $k_{12}, k_{23}, k_{34}$ whose natural lengths of extension are $l_{12}, l_{23}, l_{34}$.
Let $f_{1}, f_{2}, f_{3}, f_{4}$ denote the rightward forces on the masses, e.g.,
$f_{1} = k_{12}(x_{2} - x_{1} - l_{12})$

a) Write the 4 X 4 matrix equation relating the column vectors $f$ and $x$. Let $K$ denote the matrix in this equation.

2. Attempt at solution
I treat this as a equilibrium problem. I imagine the masses being pulled to the right and held in place. Then I find the forces neccessary to keep them in that new position. I'm not sure that this is what the problem asks for, but I don't want to just post the question :)

$f_{2} = k_{23}(x_{3} - x_{2} - l_{23}) - f_{1}$
$f_{3} = k_{34}(x_{4} - x_{3} - l_{34}) - (f_{1} + f_{2})$
$f_{4} = f_{1} + f_{2} + f_{3}$

I then expand and simplify each and expression.

I then write these equations in the form:
$
\textbf{f} = \textbf{K}\textbf{x} + \textbf{c}
$

$
\left[\begin{array}[pos]{c}
f_{1} \\
f_{2} \\
f_{3} \\
f_{4} \\
\end{array}\right]$

$
=
\left[\begin{array}[pos]{cccc}
-k_{12} & k_{12} & 0 & 0 \\
k_{12} & (-k_{12}-k_{23}) & k_{23} & 0 \\
0 & k_{23} &(-k_{23}-k_{34}) & k_{34} \\
0 & 0 &-k_{34} & k_{34}\\
\end{array}\right]$

$
\left[\begin{array}[pos]{c}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
\end{array}\right]$

$
+
\left[\begin{array}[pos]{c}
-k_{12}l_{12} \\
-k_{23}l_{23} + k_{12}l_{12} \\
-k_{34}l_{34} + k_{23}l_{23} \\
-k_{34}l_{34} \\
\end{array}\right]

$

Sorry about the formatting of matrices, I get too long string errors when I tried writing them next to eachother..
Somehow I feel that the masses should be included in the party as well...

Any suggestions are appreciated, thanks.
• Dec 1st 2009, 05:45 AM
HallsofIvy
That looks good to me!
• Dec 1st 2009, 05:59 AM
Mollier
Nothing better than that :)
Thanks!